[Guest]

A ferry service connects two small ports at either end of a lake. Two ferries depart simultaneously from each port for the other. Their paths cross 700 metres from one of the ports. They then arrive at their respective destinations, and each spends an hour disembarking and embarking passengers, before returning to their original ports. This time their paths cross 300 metres from the other port.

What is the distance between the two ports?

[Guest]

1000 m

[Guest]

1000 m

Not according to my calculation.

[Guest]

couldn't it be any distance greater than 700 meters, since we don't know the speeds of the ferries?...one could move slower than the other on the first trip, and the other move slower on the way back..must be missing something here..

[Guest]

couldn't it be any distance greater than 700 meters, since we don't know the speeds of the ferries?...one could move slower than the other on the first trip, and the other move slower on the way back..must be missing something here..

I see where you are at, James. Though I did not specifically put it into the problem, you may assume that each ferry travels with a uniform velocity throughout, though clearly the two velocities differ. Again assume that the two ferries travel in a straight line from port to port (but do not collide (!)

[bonanova]

1800 m

[bonanova]

The first meeting clearly is closer to the slower boat's home harbor: call it A.

The faster boat will have completed a greater fraction of its total trip at the second meeting.

So the distance from the second meeting to the faster boat's home harbor will be less than

the distance from the first meeting to the slower boat's home harbor.

Thus the first meeting was 700 m from A, and the second meeting was 300 m from B.

So at the first meeting, at time t₁, V_A/700 = V_B/[R-700] where R is the total distance.

At the second meeting, at time t₂, V_A/[R+300] = V_B/[2R-300]

Thus [R-700][R+300] = 700[2R-300]

This simplifies to

R[R-1800] = 0.

The 1-hour passenger time is a red herring and can be ignored.

[Guest]

I thought that you would solve this simple puzzle quickly, moderator.

It is a slight generalization of a problem I encountered in Mathematical Pie.

Ferry Problem.doc

[Guest]

A few simultaneous equations gave me

1800m

Edited August 1, 2009 by psychic_mind

[Guest]

You may have inserted the 1 hour as a red herring but it actually makes the solution incorrect. Since they arrive at the destination port at different times, they cannot possibly depart at the same time for the return trip if they each spend an hour embarking and disembarking passengers -- the faster boat must spend more than an hour or the slower boat must spend less than an hour.

[Guest]

You may have inserted the 1 hour as a red herring but it actually makes the solution incorrect. Since they arrive at the destination port at different times, they cannot possibly depart at the same time for the return trip if they each spend an hour embarking and disembarking passengers -- the faster boat must spend more than an hour or the slower boat must spend less than an hour.

Nowhere is it suggested in the question that the ferries depart for their return journeys at the same time. In fact, they can't.

[Guest]

My bad! I made a mistake in transposing my equations.